darkpool - 1 year ago 89
Python Question

# Sort each row individually between two columns

I have the following pandas dataframe:

``````column_01   column_02   value
ccc         aaa         1
bbb         ddd         34
ddd         aaa         98
``````

I need to re-organise the dataframe such that
`column_01`
contains which ever value comes first alphabetically between
`column_01`
and
`column_02`
. The output of the above example would be:

``````column_01   column_02   value
aaa         ccc         1
bbb         ddd         34
aaa         ddd         98
``````

I could obviously do this by iterating over the dataframe one row at a time, comparing
`column_01`
to
`column_02`
to see which comes first alphabetically and swapping them if necessary. The only problem with this is that the dataframe is quite big (over 1million rows), so this isn't a very efficient way to do this.

Is there a way to do this without iterating over every row individually?

You can use:

``````df[['column_01','column_02']] =
df[['column_01','column_02']].apply(lambda x: sorted(x.values), axis=1)
print (df)
column_01 column_02  value
0       aaa       ccc      1
1       bbb       ddd     34
2       aaa       ddd     98
``````

Another solutions:

``````df[['column_01','column_02']] = pd.DataFrame(np.sort(df[['column_01','column_02']].values),
index=df.index, columns=['column_01','column_02'])
``````

only with numpy array:

``````df[['column_01','column_02']] = np.sort(df[['column_01','column_02']].values)
print (df)
column_01 column_02  value
0       aaa       ccc      1
1       bbb       ddd     34
2       aaa       ddd     98
``````

Second solution is faster, because `apply` use loops:

``````df = pd.concat([df]*1000).reset_index(drop=True)
In [177]: %timeit df[['column_01','column_02']] = pd.DataFrame(np.sort(df[['column_01','column_02']].values), index=df.index, columns=['column_01','column_02'])
1000 loops, best of 3: 1.36 ms per loop

In [182]: %timeit df[['column_01','column_02']] = np.sort(df[['column_01','column_02']].values)
1000 loops, best of 3: 1.54 ms per loop

In [178]: %timeit df[['column_01','column_02']] = (df[['column_01','column_02']].apply(lambda x: sorted(x.values), axis=1))
1 loop, best of 3: 291 ms per loop
``````
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